JEE Main 2026 — Mathematics Vectors & 3D Geometry

Let the point of intersection be $(x, y, 0)$(x,y,0) since it lies on the $xy$xy-plane.

For the first line, equating the coordinates to a parameter $\lambda$λ:

$$\dfrac{x+1}{3} = \dfrac{y+a}{5} = \dfrac{0+b+1}{7} = \lambda$$3x+1​=5y+a​=70+b+1​=λ

This gives $x = \dfrac{3b+3}{7} - 1$x=73b+3​−1 and $y = \dfrac{5b+5}{7} - a$y=75b+5​−a.

For the second line, equating the coordinates to a parameter $\mu$μ:

$$\dfrac{x-2}{1} = \dfrac{y-b}{4} = \dfrac{0-2a}{7} = \mu$$1x−2​=4y−b​=70−2a​=μ

This gives $x = \dfrac{-2a}{7} + 2$x=7−2a​+2 and $y = \dfrac{-8a}{7} + b$y=7−8a​+b.

Equating the $x$x-coordinates from both lines:

$$\dfrac{3b+3}{7} - 1 = \dfrac{-2a}{7} + 2$$73b+3​−1=7−2a​+2

$$\dfrac{3b-4}{7} = \dfrac{-2a+14}{7}$$73b−4​=7−2a+14​

$$2a + 3b = 18$$2a+3b=18

Equating the $y$y-coordinates from both lines:

$$\dfrac{5b+5}{7} - a = \dfrac{-8a}{7} + b$$75b+5​−a=7−8a​+b

$$5b + 5 - 7a = -8a + 7b$$5b+5−7a=−8a+7b

$$a - 2b = -5$$a−2b=−5

Solving the two linear equations $2a + 3b = 18$2a+3b=18 and $a - 2b = -5$a−2b=−5:

From the second equation, $a = 2b - 5$a=2b−5. Substituting this into the first equation:

$$2(2b - 5) + 3b = 18$$2(2b−5)+3b=18

$$4b - 10 + 3b = 18$$4b−10+3b=18

$$7b = 28 \Rightarrow b = 4$$7b=28⇒b=4

Substituting $b = 4$b=4 into $a = 2b - 5$a=2b−5:

$$a = 2(4) - 5 = 3$$a=2(4)−5=3

Therefore, $a + b = 3 + 4 = 7$a+b=3+4=7.

Answer: $7$7